# Proving that $w^n-z=0$, for non-zero complex $z$ and positive integer $n$, has exactly $n$ distinct complex roots

I recently have been reading Basic Mathematics by Serge Lang and encountered this question

Let $$z$$ be a complex number not equal to $$0$$. Let $$n$$ be a positive integer. Show that there are $$n$$ distinct complex numbers $$w$$ such that $$w^n = z$$. Write these complex numbers in polar form. The proof given that a polynomial of degree $$n$$ has at most $$n$$ roots applies to the complex case, and thus we see that there are no other complex numbers $$w$$ such that $$w^n = z$$ other than those you have presumably written down.

I tried relating it with polynomial in form $$x^n -z = 0$$, and proceeding with that same pattern and tried to come up with solutions, but it gets quite complicated pretty quick and I wonder if it will worth to consider odd/even powers and their effect on imaginary part.

I would also love to see if there is any way to prove the $$n$$ roots thing for complex number case.

I will appreciate the help. Thanks!

The fundamental theorem of algebra guarantees the $$n$$ solutions.
Take a primitive $$n$$-th root of unity: $$\zeta=e^{2\pi i/n}$$. Then the $$n$$ solutions are $$\zeta^kz^{1/n},\,k=0,2,\dots n-1$$.
If $$z=re^{i\theta}$$, we get $$r^{1/n}e^{i(\theta+2\pi k)/n}$$.